I have a df where columns 2 and beyond are dollar amounts such as $1004.23, ($1482.40), $2423.94 etc. Similar to the example below:
> df
id desc price
Another possible solution which builds on your own attempt and takes into account that you need to transform more columns than in the example:
d[,-c(1:2)] <- lapply(d[,-c(1:2)],
function(x) as.numeric(gsub('[$,]', '', sub(")", "", sub("(", "-", x, fixed=TRUE), fixed=TRUE))))
which gives:
> d
id desc price price2
1 0 apple 1.00 -5.90
2 1 banana -2.25 2.39
3 2 grapes 1.97 -0.95
Or using a for-loop:
for(i in 3:ncol(d)){
d[[i]] <- as.numeric(gsub('[$,]', '', sub(")", "", sub("(", "-", d[[i]], fixed=TRUE), fixed=TRUE)))
}
Or using the data.table
package:
library(data.table)
cols <- names(d)[-c(1:2)]
setDT(d)[, (cols) := lapply(.SD, function(x) as.numeric(gsub('[$,]', '', sub(")", "", sub("(", "-", x, fixed=TRUE), fixed=TRUE)))),
.SDcols = cols]
Or using the dplyr
package:
library(dplyr)
d %>%
mutate_all(funs(as.numeric(gsub('[$,]', '', sub(")", "", sub("(", "-", ., fixed=TRUE), fixed=TRUE)))), -c(1:2))
which will all give you the same result.
Used data:
d <- structure(list(id = 0:2, desc = c("apple", "banana", "grapes"),
price = c("$1.00", "($2.25)", "$1.97"),
price2 = c("($5.9)", "$2.39", "($0.95)")),
.Names = c("id", "desc", "price", "price2"), class = "data.frame", row.names = c("1", "2", "3"))